Factorisation?

2h^2 - 12h + 18

= 2(h^2 - 6h + 9)
= 2(h-3)^2

Or more general:

y = ax^2 + bx + c
= x^2 + bx/a + c/a
= (x+ b/2a)^2 - (b/2a)^2 + c/a

2h^2 - 12h + 18
= 2(h^2 - 6h + 9)
= 2(h - 3)(h - 3)
= 2(h - 3)^2

The first thing to do is factor a 2 out of each term

2(h^2 - 6h + 9)

the quadratic factors as

(h-3)(h-3)

so the complete factorization is

2(h-3)(h-3)

Take out 2 common 2(h^2 - 6h + 9) = 2(h - 3)^2

h1= {-(-12)+(144-4(2)(18))^(0.5)}/2(2)
= (12+0)/4
= 3
h2={-(-12)-(144-4(2)(18))^(0.5)}/2(2)
= (12+0)/4
= 3
This equation is factorised as (h-3)*(h-3).And the two roots are real and equal which is 3

2h^2 - 12h + 18=2(h^2 - 6h + 9)=2(h-3)^2.
Does this serve your purpose?

2h^2-12h+18
=2 [h^2-6h+9]
=2[h^2-2.3.h+3^2]
=2 [(h-3)^2] ans

2h^2 - 12h + 18
2(h^2 -6h +9)
2(h^2 -3h-3h +9)
2(h(h-3)-3(h -3))
2(h-3)(h-3)
2(h-3)^2

2h^2 - 12h + 18

2h^2 - 6h - 6h + 18
2h(h - 3) - 6(h - 3)
(h-3)(2h-6)

very simple dude
and yeah you've only said factorise it ok
2h^2-12h+18=0
2h^2-6h-6h+18=0
2h(h-3)-6(h-3)=0
(2h-6) (h-3) =0
your answer is here